Biochemistry
6th Edition
ISBN: 9781305577206
Author: Reginald H. Garrett, Charles M. Grisham
Publisher: Cengage Learning
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Textbook Question
Chapter 10, Problem 20P
Answers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book.
Structural complementarity is the key to molecular recognition, a lesson learned in Chapter 1. The principle of structural complementarity is relevant to answering problems 5, 6, 7,11, 12, and 19. The quintessential example of structural complementarity in all of biology is the DNA double helix. What features of the DNA double helix exemplify structural complementarity?
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Student Guide (continued)
Record data and conclusions about the mystery food sample either below or in a lab notebook.
Step 2: Protein Test (Biuret Solution)
Gelatin
Water
[Mystery Food
(Positive Control)
(Negative Control)
Sample
pink purple
no change
no change
They mystery food sample does not contain protein because the color of the
test tube wasn't pink or purple
Color
Conclusion They mystery food sample does not contain protein because the color of the test tube
wasn't pink or purple
Step 3: Lipid Test (Sudan Red Solution)
Vegetable Oil
Water
(Positive Control)
(Negative Control)
Mystery Food
Sample
floating red
no change
floating red
the mystery food dosnt contain lipids because the test tube has floating red
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The rate data from an enzyme catalyzed reaction with and without an inhibitor present is found in the image.
Question: what is the KM and Vm and the nature of inhibition
1. Estimate the concentration of an enzyme within a living cell. Assume that:
(a): fresh tissue is 80% water and all of it is intracellular
(b): the total soluble protein represents 15% of the weight
(c): all the soluble proteins are enzymes
(d): the average molecular weight of the proteins is 150,000
(E): about 100 different enzymes are present
please help I am lost
Chapter 10 Solutions
Biochemistry
Ch. 10 - Answers to all problems are at the end of this...Ch. 10 - Answers to all problems are at the end of this...Ch. 10 - Answers to all problems are at the end of this...Ch. 10 - Answers to all problems are at the end of this...Ch. 10 - Answers to all problems are at the end of this...Ch. 10 - Answers to all problems are at the end of this...Ch. 10 - Answers to all problems are at the end of this...Ch. 10 - Prob. 8PCh. 10 - Prob. 9PCh. 10 - Prob. 10P
Ch. 10 - Answers to all problems are at the end of this...Ch. 10 - Prob. 12PCh. 10 - Prob. 13PCh. 10 - Prob. 14PCh. 10 - Answers to all problems are at the end of this...Ch. 10 - Prob. 16PCh. 10 - Prob. 17PCh. 10 - Prob. 18PCh. 10 - Prob. 19PCh. 10 - Answers to all problems are at the end of this...
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- The following data were recorded for the enzyme catalyzed conversion of S -> P Question: what would the rate be at 5.0 x 10-5 M [S] and the enzyme concentration was doubled? Also, the rate given in the table is from product accumulation after 10 minuets of reaction time. Verify these rates represent a true initial rate (less than 5% turnover). Please helparrow_forwardThe following data was obtained on isocitrate lyase from an algal species. Identify the reaction catalyzed by this enzyme, deduce the KM and Vmax , and determine the nature of the inhibition by oxaloacetate. Please helparrow_forwardIn the table below, there are sketches of four crystals made of positively-charged cations and negatively-charged anions. Rank these crystals in decreasing order of stability (or equivalently increasing order of energy). That is, select "1" below the most stable (lowest energy) crystal. Select "2" below the next most stable (next lowest energy) crystal, and so forth. A B 鹽 (Choose one) +2 C +2 +2 (Choose one) D 鹽雞 (Choose one) (Choose one)arrow_forward
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